The Simple Guide to the Lambert W Function Graph

[@grestyacademy]

For more videos on the Lambert W Function see the Gresty Academy playlist 'Lambert W Function' at kzbin.info/aero/PLlX3COjsHPVApqLukQcbj2e9rDC5sRg_t

[@bonzobobajob9851]

So clear...Making a complex subject sound easy...more please

[@grestyacademy]

thanks for the feedback!

[@leilanemag4643]

A good explanation on this subject. Thank you!

[@grestyacademy]

You're very welcome!

[@almanduku9043]

I have a question sir, if a function gives 2 results for one x, hox could it be a func.? W(-1/4) W₀ or W₁

[@grestyacademy]

Great question and in fact it isnt a function for precisely the reasons you state - so the W0 and W1 are considered as two separate branches (ie the red line and the blue line), both of which ARE functions..hope this clarifies

[@almanduku9043]

@@grestyacademy ok thanks for answer 👍🏻👍🏻

[@grestyacademy]

You are most welcome!

[@nooruddinbaqual7869]

I came across an equation. ln(x)=Wln5 Then x=ln5÷Wln5 x=1•609÷W(ln5) Then x=2•1293 Now for x to be equal to 2•1293, the value of W(ln5) has to be •7556 Could you please tell me how W(ln5) equals •7556

[@grestyacademy]

Hi there - starting with xe^x = a for some constant 'a' then by definition x=W(a) and so also by definition W(a)e^W(a)=a --- basically W(a) is simply the value of 'x' which solves xe^x = a. So given that 0e^0=0 and 0.5e^0.5~0.824 and 1e^1~2.71 and 2e^2~15 and 3e^3~60 (give or take) then W(0) = 0 and W(0.824)~0.5 and W(2.71)~1 and W(15)~2 and W(60)~3. Now ln(5)~1.61 (from a calculator) so W(ln 5)= W(~1.61)~0.75 interpolating between the values we can calculate easily W(0.824) and W(2.71). Hope this clarifies. Incidentally, letting y=ln(x) in your equation (to make the explanation easier) we have y=W(ln5) so from the definition ye^y=ln(5), ie ln(x)e^(lnx)=ln(5) and as e^lnx=x we have xln(x)=ln(5) ie ln(x^x)=ln5 ie x^x = 5 so the solution 'x' to the power of itself is 5 :)

[@laabida]

Is he confusing equations, functions and relations?

[@grestyacademy]

Hi there - if you think there is an error in the video feel free to detail the error, what it should say, and the time in the video the error occurred (ie at 5:35 or whatever). We check all our videos before publishing but there is always a possibility an error has somehow found its way through undetected. Thanks

[@MathTidbits]

is there a handheld scientific calculator like casio that will give lambert function values ?

[@grestyacademy]

I always use www.wolframalpha.com/ inputting for example LambertW[0,2] or LambertW[-1,1/4] etc

[@Rai_Te]

With almost all actual programmable calculators that include a solve() function in their library, this is relatively easy to achieve: Given that V is the input value that you put into the W() function, you solve the equation x * e^x - V = 0 With the startvalue of 1 you will get the solution from the 0-branch and with a startvalue of -1.5 you will get the solution from the -1-branch (if there is one ... if there is none, you will get the same solution as the 0-branch). I takes only about 10 lines of code in my casio 9680g3 ... unfortunately I cannot post them here ... somehow code in comments get the comment deleted automatically.

[@grestyacademy]

Useful comment thanks for posting

[@richardcarnegie777]

Nice overview (“mirror”) before going into the “weeds”…

[@grestyacademy]

Thanks😎

[@matthewstoicism1485]

if you had to calculate W(5) by hand; how would you go about it?

[@grestyacademy]

Hi there - interesting question. There are lots of online calculators around but if you consider 'excel' calculating by hand you could 'build' a Lambert W function calculator into that - for real 'hand' method I would suggest Newton Raphson iteration. Good luck!

[@matthewstoicism1485]

@@grestyacademy in my mind's eye I can see how to undo multiplication by way of division: I can see how to square root a square . . . but what is happening when I w(5)? are you saying that it can't be done by hand? if it can't be done by hand then are you sure it's real? I can just graph it on Desmos and just ''tap'' the answer: however, what would they do before calculators? Imagin telling a machinist to cut something to a length of W(5)? or perhaps each point of datum is just a reflection across the line x=y and therefore there is no process?

[@grestyacademy]

Hi there, I am not saying it can't be done by hand - it can. Personally I consider it similar to logarithms in that they can be calculated without calculators (via log tables in previous times) but normally they don't need to be. So W(5) is just as real as log (5). W(5) = 1.3267 (approx) and all that means is that 1.3267 is the solution to xe^x = 5. So one way of doing it by hand would be to put random values of x into xe^x and then narrow them down (a basic Newton Raphson). So for example to find w(5) we could put x=1 into xe^x and that gives us 'e' (2.71 approx) so lets put x=2 and that gives us 2e^2 (14.7 approx) - so we know w(5) is between 1 and 2, so lets try 1.5 which gives us 1.5e^1.5 = 6.72 approx so we know w(5) is between 1 and 1.5 etc etc. So yes a machinist asked to cut something to length w(5) could (without a calculator) look up the value in a 'lambert W function table' (similar to a log table) and get the measurement 1.3267. Hope this helps

[@matthewstoicism1485]

@@grestyacademy . . . let's try these two: if someone were to ask me ''what is the sine of 89.99999999 degrees?'', I would say to them "I don't know, but it must be close to one"; if theta is less than 10 degrees and I know the length of the Tangent line then I know that I can just use its length as an approx for the sine function, yet I would not swap them if the theta was 80 degrees. I can have these thoughts because I know what is really going on --the why of it--, yet if someone were to ask me in earnest ''what is W(5) approx?'', my thoughts would be empty. like the log tables sometimes one must play collect the dots until a wiggle graph appears, and hope that the wiggle can be reflected across the line x=y. thanks for the help.

[@grestyacademy]

Hi there I get your point. For me in a similar way that I can estimate in my head that log (to base 10) of 50 for example is somewhere slightly north of 1.5 because I know that log 10=1 and log 100 = 2, I can approximate W(50) for example in my head to somewhere in the region of 2.75 (its actually 2.86) because I know that w(e)=1 (e~2.71) and w(2e^2)=2 (2e^2~2*2.71*2,71~15 give or take)and w(3e^3)=3 (3e^3~3*2.71*2,71*2.71~ 60 give or take) etc ...so very very roughly as pointers w(2.71)~1 and w(15)~2 and w(60)~3...From that I can estimate that w(10)~1.75 , w(30)~2.4 etc etc which are not that far from their true values. Hope this helps

[@brendangolledge8312]

Somehow I feel like this is cheap. I suppose you can just define a new function every time you don't know the answer? Suppose I define a "Gangsta function" such that G(y) = x when y = x*cos(x). So then for 1 = x*cos(x), x = G(1). This doesn't actually give me any new information, because if I wanted the numerical answer, I'd still have to use some kind of calculator. It looks like for any arbitrary "y", G(y) has infinitely many solutions. Then this makes me feel like much of algebra is a cheap trick, where I can write all manner of values using sin, cos, e, ln, etc, but I have no ability to accurately write down their numerical value in decimal form. I suppose if you're trying to get a numerical answer, all of these functions are useless in the absence of some method for approximating the numerical value. Do calculators use Taylor series to calculate the values of these functions? Then they'd be able to calculate anything using only addition, subtraction, multiplication, and division.

[@grestyacademy]

valid point!

[@almanduku9043]

Calculators do it partially

[@TelepathShield]

Well the same is true for logarithms for example as being the inverse of exponentiation. It just makes the writing easier. And you could always use newton’s method to find an output

[@donfzic7471]

La fonction de LAMERT résosud bon nombre d'équations en matémathiques et sciences physiques Exemple : x^4 = 4.x. Equation facile. Equation plus compliquée : 4^x = 4.x. Non facile à résoudre. Utilser la fonction de 'LAMBERT W' dans ce cas.

[@grestyacademy]

Hi - we have a video on that - check it out at kzbin.info/www/bejne/epi4kJuBaMqShsk (entitled 'How to Solve 3^x = 2x + 2')

[@aliciaaquino7909]

ano yung lambert.?

[@grestyacademy]

named after en.wikipedia.org/wiki/Johann_Heinrich_Lambert

[@mecha37000fighter]

Thank you, this was very helpful in understanding the subject🫡

[@grestyacademy]

Glad to be of service - thanks for the feedback. We have a few Lambert W Function videos up on our channel if you are interested - check out the playlist 'Lambert W Function' at kzbin.info/aero/PLlX3COjsHPVApqLukQcbj2e9rDC5sRg_t